
/**
Description

How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.

Input

The input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c will contain exactly three digits.
Output

For each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples.
Sample Input

1.00
3.71
0.04
5.19
0.00
Sample Output

3 card(s)
61 card(s)
1 card(s)
273 card(s)
 */

import java.util.Scanner;

public class P1003 {
	
	static int K=276;
	static double[] buf = new double[K+1];
	
	public static void main(String[] args){

		buf[0]=0;
		for(int i=1;i<=K;i++){
			buf[i]=buf[i-1]+((double)1)/(i+1);
		}
		
		Scanner cin=new Scanner(System.in);
		String line;
		double d;
		int index=0;
        while(!(line=cin.nextLine()).equals("0.00")){
        	d = Double.parseDouble(line);
//        	int index = bSearch(d,1,276);
        	for(int i=1;i<=276;i++){
        		if(buf[i]>=d){
        			index=i;
        			break;
        		}
        	}
        	System.out.println(index+" card(s)");
        	
        }
	}
//	
//	public static int bSearch(double value,int left, int right){
//		
//		if(left==right){
//			if(buf[left]>=value)
//				return left;
//			else
//				return left+1;
//		}
//		
//		int mid=(left+right)/2;
//		if(buf[mid]==value){
//			return mid;
//		}else{
//			if(buf[mid]<value){
//				return bSearch(value,mid+1,right);
//			}else{
//				return bSearch(value,left,mid-1);
//			}
//		}
//	}
}
